Prove that the product of two lower-triangular matrices is lower-triangular.

Let \(A = (a_{ij})\) and \(B = (b_{ij})\) be two \(n \times n\) lower-triangular matrices. By definition, \(A\) being lower-triangular means \(a_{ij} = 0\) for all \(i < j\) and \(B\) being lower-triangular means \(b_{ij} = 0\) for all \(i < j\).

Let \(C\) be their product: \(C = AB = (c_{ij})\), where \(c_{ij} = \sum\limits_{k=1}^n a_{ik}b_{kj}\). We will show that this product is also lower-triangular: \(c_{ij} = 0\) for all \(i < j\):

\[\begin{split} c_{ij} &= \sum\limits_{k=1}^n a_{ik}b_{kj} \\ &= \sum\limits_{k=1}^{j-1} a_{ik} b_{kj} + \sum\limits_{k=j}^n a_{ik} b_{kj} \\ &= \sum\limits_{k=1}^{j-1} a_{ik} 0 + \sum\limits_{k=j}^n 0 b_{kj} \\ &= \sum\limits_{k=1}^{j-1} 0 + \sum\limits_{k=j}^n 0 \\ &= 0 + 0 \\ &= 0 \end{split}\]